Answer:
(a) The value of z is 0.90.
(b) The value of z is 0.30.
(c) The value of z is 0.86.
(d) The value of z is 1.25.
Explanation:
If
, then
, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is,
.
(a)
Compute the value of z for P (Z < z) = 0.8164 as follows:
P (Z < z) = 0.8164
The probability is approximately true for z = 0.90.
Thus, the value of z is 0.90.
(b)
Compute the value of z for P (Z < z) = 0.6192 as follows:
P (Z < z) = 0.6192
The probability is approximately true for z = 0.30.
Thus, the value of z is 0.30.
(c)
Compute the value of z for P (Z > z) = 0.1941 as follows:
P (Z > z) = 0.1941
P (Z < z) = 1 - 0.1941
= 0.8059
The probability is approximately true for z = 0.86.
Thus, the value of z is 0.86.
(d)
Compute the value of z for P (Z > z) = 0.1055 as follows:
P (Z > z) = 0.1055
P (Z < z) = 1 - 0.1055
= 0.8945
The probability is approximately true for z = 1.25.
Thus, the value of z is 1.25.